A data consists of n observations

{x_1},{x | vedclass

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Question

${\sum {\left( {{x_i} + 1} \right)} ^2} = 9n\,\,\,\,\,\,....\left( 1 \right)$

${\sum {\left( {{x_i} - 1} \right)} ^2} = 5n\,\,\,\,\,\,....\left( 2

Vedclass · Accepted Answer

$\sqrt 5$

Vedclass · Answer

${\sum {\left( {{x_i} + 1} \right)} ^2} = 9n\,\,\,\,\,\,....\left( 1 \right)$

${\sum {\left( {{x_i} - 1} \right)} ^2} = 5n\,\,\,\,\,\,....\left( 2 \right)$

$\left( 1 \right) + \left( 2 \right) \Rightarrow \sum {\left( {x_i^2 + 1} \right)}  = 7n$

$ \Rightarrow \frac{{\sum {x_i^2} }}{n} = 6$

$\left( 1 \right).\left( 2 \right) \Rightarrow 4\sum {{x_i}}  = 4n$

$ \Rightarrow \sum {{x_i} = n} $

$ \Rightarrow \frac{{\sum {{x_i}} }}{n} = 1$

$ \Rightarrow $ variance $=6-1=5$

$ \Rightarrow $standard diviation $ = \sqrt 5 $

	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

A data consists of $n$ observations

${x_1},{x_2},......,{x_n}.$ If $\sum\limits_{i - 1}^n {{{({x_i} + 1)}^2}} = 9n$ and $\sum\limits_{i - 1}^n {{{({x_i} - 1)}^2}} = 5n,$ then the standard deviation of this data is

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A data consists of $n$ observations ${x_1},{x_2},......,{x_n}.$ If $\sum\limits_{i - 1}^n {{{({x_i} + 1)}^2}} = 9n$ and $\sum\limits_{i - 1}^n {{{({x_i} - 1)}^2}} = 5n,$ then the standard deviation of this data is